3.504 \(\int \tan ^3(c+d x) \sqrt{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=159 \[ -\frac{4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac{2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac{2 \sqrt{a+b \tan (c+d x)}}{d}+\frac{\sqrt{a-i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{\sqrt{a+i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d} \]

[Out]

(Sqrt[a - I*b]*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d + (Sqrt[a + I*b]*ArcTanh[Sqrt[a + b*Tan[c +
d*x]]/Sqrt[a + I*b]])/d - (2*Sqrt[a + b*Tan[c + d*x]])/d - (4*a*(a + b*Tan[c + d*x])^(3/2))/(15*b^2*d) + (2*Ta
n[c + d*x]*(a + b*Tan[c + d*x])^(3/2))/(5*b*d)

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Rubi [A]  time = 0.333801, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {3566, 3630, 12, 3528, 3539, 3537, 63, 208} \[ -\frac{4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac{2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac{2 \sqrt{a+b \tan (c+d x)}}{d}+\frac{\sqrt{a-i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{\sqrt{a+i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3*Sqrt[a + b*Tan[c + d*x]],x]

[Out]

(Sqrt[a - I*b]*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d + (Sqrt[a + I*b]*ArcTanh[Sqrt[a + b*Tan[c +
d*x]]/Sqrt[a + I*b]])/d - (2*Sqrt[a + b*Tan[c + d*x]])/d - (4*a*(a + b*Tan[c + d*x])^(3/2))/(15*b^2*d) + (2*Ta
n[c + d*x]*(a + b*Tan[c + d*x])^(3/2))/(5*b*d)

Rule 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3539

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \tan ^3(c+d x) \sqrt{a+b \tan (c+d x)} \, dx &=\frac{2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac{2 \int \sqrt{a+b \tan (c+d x)} \left (-a-\frac{5}{2} b \tan (c+d x)-a \tan ^2(c+d x)\right ) \, dx}{5 b}\\ &=-\frac{4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac{2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac{2 \int -\frac{5}{2} b \tan (c+d x) \sqrt{a+b \tan (c+d x)} \, dx}{5 b}\\ &=-\frac{4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac{2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\int \tan (c+d x) \sqrt{a+b \tan (c+d x)} \, dx\\ &=-\frac{2 \sqrt{a+b \tan (c+d x)}}{d}-\frac{4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac{2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\int \frac{-b+a \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 \sqrt{a+b \tan (c+d x)}}{d}-\frac{4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac{2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac{1}{2} (-i a-b) \int \frac{1+i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx-\frac{1}{2} (i a-b) \int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 \sqrt{a+b \tan (c+d x)}}{d}-\frac{4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac{2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac{(a-i b) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 d}-\frac{(a+i b) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d}\\ &=-\frac{2 \sqrt{a+b \tan (c+d x)}}{d}-\frac{4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac{2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac{(i a-b) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i a}{b}-\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}-\frac{(i a+b) \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i a}{b}+\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}\\ &=\frac{\sqrt{a-i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d}+\frac{\sqrt{a+i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d}-\frac{2 \sqrt{a+b \tan (c+d x)}}{d}-\frac{4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac{2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}\\ \end{align*}

Mathematica [A]  time = 1.05093, size = 140, normalized size = 0.88 \[ \frac{\frac{2 \sqrt{a+b \tan (c+d x)} \left (-2 a^2+a b \tan (c+d x)+3 b^2 \tan ^2(c+d x)-15 b^2\right )}{b^2}+15 \sqrt{a-i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )+15 \sqrt{a+i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3*Sqrt[a + b*Tan[c + d*x]],x]

[Out]

(15*Sqrt[a - I*b]*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] + 15*Sqrt[a + I*b]*ArcTanh[Sqrt[a + b*Tan[c
+ d*x]]/Sqrt[a + I*b]] + (2*Sqrt[a + b*Tan[c + d*x]]*(-2*a^2 - 15*b^2 + a*b*Tan[c + d*x] + 3*b^2*Tan[c + d*x]^
2))/b^2)/(15*d)

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Maple [B]  time = 0.069, size = 519, normalized size = 3.3 \begin{align*}{\frac{2}{5\,{b}^{2}d} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{2\,a}{3\,{b}^{2}d} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-2\,{\frac{\sqrt{a+b\tan \left ( dx+c \right ) }}{d}}+{\frac{1}{4\,d}\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}\ln \left ( b\tan \left ( dx+c \right ) +a+\sqrt{a+b\tan \left ( dx+c \right ) }\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}+\sqrt{{a}^{2}+{b}^{2}} \right ) }-{\frac{a}{d}\arctan \left ({ \left ( 2\,\sqrt{a+b\tan \left ( dx+c \right ) }+\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}}+{\frac{1}{d}\arctan \left ({ \left ( 2\,\sqrt{a+b\tan \left ( dx+c \right ) }+\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \right ) \sqrt{{a}^{2}+{b}^{2}}{\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}}-{\frac{1}{4\,d}\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}\ln \left ( \sqrt{a+b\tan \left ( dx+c \right ) }\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}-b\tan \left ( dx+c \right ) -a-\sqrt{{a}^{2}+{b}^{2}} \right ) }+{\frac{a}{d}\arctan \left ({ \left ( \sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}-2\,\sqrt{a+b\tan \left ( dx+c \right ) } \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}}-{\frac{1}{d}\arctan \left ({ \left ( \sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}-2\,\sqrt{a+b\tan \left ( dx+c \right ) } \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \right ) \sqrt{{a}^{2}+{b}^{2}}{\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^3,x)

[Out]

2/5/d/b^2*(a+b*tan(d*x+c))^(5/2)-2/3*a*(a+b*tan(d*x+c))^(3/2)/b^2/d-2*(a+b*tan(d*x+c))^(1/2)/d+1/4/d*(2*(a^2+b
^2)^(1/2)+2*a)^(1/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))-1
/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^
(1/2)-2*a)^(1/2))*a+1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)
^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*(a^2+b^2)^(1/2)-1/4/d*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*ln((a+b*tan(d*x+c))
^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))+1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(
((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a-1/d/(2*(a^2+b^2)^(1/
2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*(
a^2+b^2)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^3,x, algorithm="maxima")

[Out]

integrate(sqrt(b*tan(d*x + c) + a)*tan(d*x + c)^3, x)

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Fricas [B]  time = 2.51524, size = 4277, normalized size = 26.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^3,x, algorithm="fricas")

[Out]

-1/60*(60*sqrt(2)*b^2*d^5*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*sqrt(b^2/d^4)*((a^2 + b^2)/d^4)
^(3/4)*arctan(-((a^2 + b^2)*d^4*sqrt(b^2/d^4)*sqrt((a^2 + b^2)/d^4) + (a^3 + a*b^2)*d^2*sqrt(b^2/d^4) + sqrt(2
)*(d^7*sqrt(b^2/d^4)*sqrt((a^2 + b^2)/d^4) + a*d^5*sqrt(b^2/d^4))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d
*x + c))*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*((a^2 + b^2)/d^4)^(3/4) - sqrt(2)*(d^7*sqrt(b^2/
d^4)*sqrt((a^2 + b^2)/d^4) + a*d^5*sqrt(b^2/d^4))*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*sqrt(((
a^2 + b^2)*d^2*sqrt((a^2 + b^2)/d^4)*cos(d*x + c) + sqrt(2)*(a*d^3*sqrt((a^2 + b^2)/d^4)*cos(d*x + c) + (a^2 +
b^2)*d*cos(d*x + c))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4)
- a^2 - b^2)/b^2)*((a^2 + b^2)/d^4)^(1/4) + (a^3 + a*b^2)*cos(d*x + c) + (a^2*b + b^3)*sin(d*x + c))/((a^2 + b
^2)*cos(d*x + c)))*((a^2 + b^2)/d^4)^(3/4))/(a^2*b^2 + b^4))*cos(d*x + c)^2 + 60*sqrt(2)*b^2*d^5*sqrt(-(a*d^2*
sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*sqrt(b^2/d^4)*((a^2 + b^2)/d^4)^(3/4)*arctan(((a^2 + b^2)*d^4*sqrt(b^2
/d^4)*sqrt((a^2 + b^2)/d^4) + (a^3 + a*b^2)*d^2*sqrt(b^2/d^4) - sqrt(2)*(d^7*sqrt(b^2/d^4)*sqrt((a^2 + b^2)/d^
4) + a*d^5*sqrt(b^2/d^4))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d
^4) - a^2 - b^2)/b^2)*((a^2 + b^2)/d^4)^(3/4) + sqrt(2)*(d^7*sqrt(b^2/d^4)*sqrt((a^2 + b^2)/d^4) + a*d^5*sqrt(
b^2/d^4))*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*sqrt(((a^2 + b^2)*d^2*sqrt((a^2 + b^2)/d^4)*cos
(d*x + c) - sqrt(2)*(a*d^3*sqrt((a^2 + b^2)/d^4)*cos(d*x + c) + (a^2 + b^2)*d*cos(d*x + c))*sqrt((a*cos(d*x +
c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*((a^2 + b^2)/d^4)^(1/4
) + (a^3 + a*b^2)*cos(d*x + c) + (a^2*b + b^3)*sin(d*x + c))/((a^2 + b^2)*cos(d*x + c)))*((a^2 + b^2)/d^4)^(3/
4))/(a^2*b^2 + b^4))*cos(d*x + c)^2 - 15*sqrt(2)*(a*b^2*d^3*sqrt((a^2 + b^2)/d^4)*cos(d*x + c)^2 + (a^2*b^2 +
b^4)*d*cos(d*x + c)^2)*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*((a^2 + b^2)/d^4)^(1/4)*log(((a^2
+ b^2)*d^2*sqrt((a^2 + b^2)/d^4)*cos(d*x + c) + sqrt(2)*(a*d^3*sqrt((a^2 + b^2)/d^4)*cos(d*x + c) + (a^2 + b^2
)*d*cos(d*x + c))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^
2 - b^2)/b^2)*((a^2 + b^2)/d^4)^(1/4) + (a^3 + a*b^2)*cos(d*x + c) + (a^2*b + b^3)*sin(d*x + c))/((a^2 + b^2)*
cos(d*x + c))) + 15*sqrt(2)*(a*b^2*d^3*sqrt((a^2 + b^2)/d^4)*cos(d*x + c)^2 + (a^2*b^2 + b^4)*d*cos(d*x + c)^2
)*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*((a^2 + b^2)/d^4)^(1/4)*log(((a^2 + b^2)*d^2*sqrt((a^2
+ b^2)/d^4)*cos(d*x + c) - sqrt(2)*(a*d^3*sqrt((a^2 + b^2)/d^4)*cos(d*x + c) + (a^2 + b^2)*d*cos(d*x + c))*sqr
t((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*((a^2 +
b^2)/d^4)^(1/4) + (a^3 + a*b^2)*cos(d*x + c) + (a^2*b + b^3)*sin(d*x + c))/((a^2 + b^2)*cos(d*x + c))) - 8*(3
*a^2*b^2 + 3*b^4 - 2*(a^4 + 10*a^2*b^2 + 9*b^4)*cos(d*x + c)^2 + (a^3*b + a*b^3)*cos(d*x + c)*sin(d*x + c))*sq
rt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c)))/((a^2*b^2 + b^4)*d*cos(d*x + c)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \tan{\left (c + d x \right )}} \tan ^{3}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**(1/2)*tan(d*x+c)**3,x)

[Out]

Integral(sqrt(a + b*tan(c + d*x))*tan(c + d*x)**3, x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^3,x, algorithm="giac")

[Out]

Timed out